Problem: On the way to school, Karen became fixated on the puzzle game on her phone!

<image>

The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0.

One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.

To win the level, after all the moves, the number in the cell at the i-th row and j-th column should be equal to gi, j.

Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!

Input

The first line of input contains two integers, n and m (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.

The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains gi, j (0 ≤ gi, j ≤ 500).

Output

If there is an error and it is actually not possible to beat the level, output a single integer -1.

Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.

The next k lines should each contain one of the following, describing the moves in the order they must be done:

  * row x, (1 ≤ x ≤ n) describing a move of the form "choose the x-th row". 
  * col x, (1 ≤ x ≤ m) describing a move of the form "choose the x-th column". 



If there are multiple optimal solutions, output any one of them.

Examples

Input

3 5
2 2 2 3 2
0 0 0 1 0
1 1 1 2 1


Output

4
row 1
row 1
col 4
row 3


Input

3 3
0 0 0
0 1 0
0 0 0


Output

-1


Input

3 3
1 1 1
1 1 1
1 1 1


Output

3
row 1
row 2
row 3

Note

In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:

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In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.

In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:

<image>

Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int N = 110;
const int INF = 1e9;
int tablero[N][N];
int n, m;
int columna[N];
int fila[N];
int sumaFila;
int sumaColumna;
void calculaFila(int i, int *res) {
  int menor = INF;
  for (int j = 0; j < m; j++) {
    menor = min(menor, tablero[i][j]);
  }
  sumaFila = 0;
  for (int j = 0; j < m; j++) {
    res[j] = tablero[i][j] - menor;
    sumaFila += res[j];
  }
}
void calculaColumna(int j, int *res) {
  int menor = INF;
  for (int i = 0; i < n; i++) {
    menor = min(menor, tablero[i][j]);
  }
  sumaColumna = 0;
  for (int i = 0; i < n; i++) {
    res[i] = tablero[i][j] - menor;
    sumaColumna += res[i];
  }
}
bool resuelve() {
  int total = 0;
  scanf("%d %d", &n, &m);
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
      scanf("%d", &tablero[i][j]);
      total += tablero[i][j];
    }
  }
  calculaFila(0, fila);
  calculaColumna(0, columna);
  int operaciones = sumaFila + sumaColumna;
  int usados = sumaFila * n + sumaColumna * m;
  int faltante = total - usados;
  int extra = faltante / (n * m);
  if (n < m) {
    operaciones += n * extra;
    for (int i = 0; i < n; i++) {
      columna[i] += extra;
    }
  } else {
    operaciones += m * extra;
    for (int j = 0; j < m; j++) {
      fila[j] += extra;
    }
  }
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
      if (i > 0 &&
          tablero[i][j] - tablero[i - 1][j] != columna[i] - columna[i - 1]) {
        return false;
      }
      if (j > 0 && tablero[i][j] - tablero[i][j - 1] != fila[j] - fila[j - 1]) {
        return false;
      }
      if (tablero[i][j] != fila[j] + columna[i]) {
        return false;
      }
    }
  }
  printf("%d\n", operaciones);
  for (int i = 0; i < n; i++) {
    for (int x = 0; x < columna[i]; x++) {
      printf("row %d\n", i + 1);
    }
  }
  for (int j = 0; j < m; j++) {
    for (int x = 0; x < fila[j]; x++) {
      printf("col %d\n", j + 1);
    }
  }
  return true;
}
int main() {
  if (!resuelve()) {
    printf("-1\n");
  }
}